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AMC 10 Preparation

AMC 10 Preparation. SDMC Euler class Instructor: David Balmin dbalmin@gmail.com. Problem 1 (Number Theory). Prove that, if the sum of two natural numbers equals 30030, then the product of the same two numbers is not divisible by 30030. Problem 1 (Number Theory).

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AMC 10 Preparation

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  1. AMC 10 Preparation SDMC Euler class Instructor: David Balmin dbalmin@gmail.com

  2. Problem 1 (Number Theory) Prove that, if the sum of two natural numbers equals 30030, then the product of the same two numbers is not divisible by 30030.

  3. Problem 1 (Number Theory) • Prime factorization of 30030: 30030 = 2*3*5*7*11*13

  4. Problem 1 (Number Theory) • Let x and (30030 - x) be two natural numbers whose sum equals 30030. So, 30030 – x > 0. • If x*(30030 - x) is divisible by 30030, then: x*(30030 - x) = 30030 *k, where k is a natural number.

  5. Problem 1 (Number Theory) • x*(30030 - x) = 30030 *k • x^2 = 30030(x – k) • 30030 = 2*3*5*7*11*13 • 2*3*5*7*11*13 | x^2 • 2*3*5*7*11*13 | x • 30030 | x • x >= 30030, that contradicts (30030 – x) > 0.

  6. AMC 10A 2012 Problem #21

  7. AMC 10A 2012 Problem #21

  8. AMC 10A 2012 Problem #21

  9. AMC 10A 2012 Problem #21 • CB = sqrt(1 + 4) = sqrt(5) • GF = sqrt(5) /2 • AD = 3 • HG = 3/2 • A(EFGH) = (3/4)* sqrt(5) The answer: (C).

  10. AMC 12A 2012 Problem #22

  11. AMC 12A 2012 Problem #22 A straight line segment connects every two midpoints of the edges within each face of a cube. A collection of distinct planes is such that the intersection of the union of all planes with the surface of the cube consists of all such segments. What is the difference between the maximum and the minimum numbers of planes that such collection can have?

  12. AMC 12A 2012 Problem #22 (A) 8 (B) 12 (C) 20 (D) 23 (E) 24

  13. AMC 12A 2012 Problem #22

  14. AMC 12A 2012 Problem #22 • 6 x 6 = 36 segments • 2 x 6 = 12 “edge-parallel” segments • 4 x 6 = 24 “diagonal” segments

  15. AMC 12A 2012 Problem #22 • Case I: a plane goes through segments 2-2 • Case II: segments 1-1 or 3-3 • Case III: segments 1-2, 2-1, 3-2, or 2-3 • Case IV: segments 1-3 or 3-1

  16. AMC 12A 2012 Problem #22 • Case I: 3 planes that cover 3 x 4 = 12 “edge-parallel” segments cut off 3 parallelepipeds:

  17. AMC 12A 2012 Problem #22 • Case II: 8 planes that cover 8 x 3 = 24 “diagonal” segments cut off 8 tetrahedra:

  18. AMC 12A 2012 Problem #22 • Case III: 12 planes that cover 12 x 2 = 24 “diagonal” segments and (12 x 2)/2 = 12 “edge-parallel” segments cut off 12 prisms:

  19. AMC 12A 2012 Problem #22 • Case IV: 4 planes that cover 6 x 4 = 24 “diagonal” segments cut off 4 heptahedra:

  20. AMC 12A 2012 Problem #22

  21. AMC 12A 2012 Problem #22 A plane is a graph of linear function z = ax + by + c.

  22. AMC 12A 2012 Problem #22 The first 3 midpoints (0,1,0), (1,0,0), (2,0,1) (or the first 2 “diagonal” segments) define the plane to which they belong: z = x + y - 1

  23. AMC 12A 2012 Problem #22 Simply check that 3 other midpoints (2,1,2), (1,2,2), (0,2,1) (or 4 other “diagonal” segments) satisfy equation z = x + y – 1 and belong to the same plane.

  24. AMC 12A 2012 Problem #22 • Case I: 3 planes (cover 12 "edge-parallel" segments) • Case II: 8 planes (cover 24 “diagonal” segments) • Case III: 12 planes (cover 24 “diagonal” segments and 12 "edge-parallel" segments) • Case IV: 4 planes (cover 24 “diagonal” segments)

  25. AMC 12A 2012 Problem #22 • Minimal collection of planes consists of 3 + 4 = 7 planes (types I and IV). • Maximal collection of planes consists of 3 + 4 + 8 + 12 = 27 planes (types I, II, III, and IV). • The answer: 27 – 7 = 20. (C)

  26. AMC 12 2001, Problem #11

  27. AMC 12 2001, Problem #11 • The drawings can end after drawing #2, #3, or maximum #4. • The “lucky” outcome: 2 white chips are drawn in any 2 drawings with numbers #1 through #4. • The “universal” outcome: 2 white chips are drawn in any 2 drawings with numbers #1 through #5.

  28. AMC 12 2001, Problem #11 • Number of events that result in the “lucky” outcome: C(4, 2) = (4*3)/2 = 6. • Number of all possible events that result in the “universal” outcome: C(5, 2) = (5*4)/2 = 10. • Probability of the “lucky” outcome: 6/10 = 3/5. • Answer: (D)

  29. AMC 12 2001, Problem #11 • Number of events that result in the “lucky” outcome: 3 ways for one red chip to be left after 4 drawings. • Number of all possible events that result in the “universal” outcome: 3 + 2 = 5 ways for one red chip or one white chip to be left after 4 drawings. • Probability of the “lucky” outcome: 3/5.

  30. AMC 10 2012, Problem #23

  31. AMC 10 2012, Problem #23 • Case 1: Each member has 1 friend • Case 2: Each member has 2 friends • Case 3: Each member has 3 friends • Case 4: Each member has 4 friends • Case 1 – mirror reflection – Case 4 • Case 2 – mirror reflection – Case 3

  32. AMC 10 2012, Problem #23 • Case 4 consists of arrangements opposite to case-1 arrangements: Each member in case-1 arrangement has 1 friend and 4 “not-friends”. Those 4 “not-friends” are friends of the same member in the corresponding case-4 arrangement. So, the number of case-4 arrangements is equal to the number of case-1 arrangements.

  33. AMC 10 2012, Problem #23 • Case 3 consists of arrangements opposite to case-2 arrangements: Each member in case-2 arrangement has 2 friends and 3 “not-friends”. Those 3 “not-friends” are friends of the same member in the corresponding case-3 arrangement. So, the number of case-3 arrangements is equal to the number of case-2 arrangements.

  34. AMC 10 2012, Problem #23 • We need to calculate the total number of case-1 and case-2 arrangements and double this total to answer the question of the problem.

  35. AMC 10 2012, Problem #23 Case 1: • First member has 5 ways to select 1 friend. • Within each such selection, first remaining member has 3 ways to select 1 friend. • Within each of these 5 x 3 = 15 selections, the remaining two members must be friends (no choices are left).

  36. AMC 10 2012, Problem #23 Case 2: • First member has C(5,2) = 10 ways to select 2 friends. • Within each such selection, there are 2 sub-cases 2.1 and 2.2.

  37. AMC 10 2012, Problem #23 Sub-case 2.1: • Two members selected by the first member are also friends. • 1 arrangement, because no more choices are left (3 remaining members must all be friends). • Number of arrangements in sub-case 2.1: 1.

  38. AMC 10 2012, Problem #23 Sub-case 2.2: • One of two members selected by the first member has 3 choices to select 1 more friend. • Once this choice has been made, the second of two members selected by the first member has 2 ways to select one of 2 remaining members. Then, no more choices are left. • Number of arrangements in sub-case 2.2: 6.

  39. AMC 10 2012, Problem #23 • Total number of arrangements in sub-cases 2.1 and 2.2: 1 + 6 = 7. • Total number of arrangements in case 2: 10 x 7 = 70. • Total number of arrangements in cases 1 and 2: 15 + 70 = 85. • Total number of arrangements in cases 1, 2, 3 and 4: 85 x 2 = 170. The answer: (B).

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